Can a Suitable Small Angle Formula Solve This Summation Problem?

[ghostyc]

Homework Statement

$$S_N(x)= \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin ((2 n-1)x)}{2 n-1}$$

By considering a suitable small angle formula show that the value of the sum at this point is

$$S_N \Big( \frac{\pi}{2 N} \Big)=\frac{2}{\pi} \int_0^{\pi} \frac{\sin (\mu)}{\mu} \; d{\mu}$$

Homework Equations

i have no idea how to get the suitable small angle formula working with this problem

The Attempt at a Solution

I have shown that

$$S_N(x)$$

can be written as

$$S_N(x)=\frac{2}{\pi} \int_0^{x} \frac{\sin (2 N t)}{\sin (t) } \; d{t}$$

my guess for suitable small angle formula is

$$\sin (x) \approx x$$ when x is small


Thank you for any help

 

[ghostyc]

anyone got any ideas? :P